Solve {left(2sqrt{2}-3sqrt{3}right)}^2

$${ \left(2 \sqrt{ 2 } -3 \sqrt{ 3 } \right) }^{ 2 }$$

Evaluate

$$35-12\sqrt{6}\approx 5.606123087$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2\sqrt{2}-3\sqrt{3}\right)^{2}.

$$4\left(\sqrt{2}\right)^{2}-12\sqrt{2}\sqrt{3}+9\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{2} is 2.

$$4\times 2-12\sqrt{2}\sqrt{3}+9\left(\sqrt{3}\right)^{2}$$

Multiply 4 and 2 to get 8.

$$8-12\sqrt{2}\sqrt{3}+9\left(\sqrt{3}\right)^{2}$$

To multiply \sqrt{2} and \sqrt{3}, multiply the numbers under the square root.

$$8-12\sqrt{6}+9\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$8-12\sqrt{6}+9\times 3$$

Multiply 9 and 3 to get 27.

$$8-12\sqrt{6}+27$$

Add 8 and 27 to get 35.

$$35-12\sqrt{6}$$

Expand

$$35-12\sqrt{6}$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2\sqrt{2}-3\sqrt{3}\right)^{2}.

$$4\left(\sqrt{2}\right)^{2}-12\sqrt{2}\sqrt{3}+9\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{2} is 2.

$$4\times 2-12\sqrt{2}\sqrt{3}+9\left(\sqrt{3}\right)^{2}$$

Multiply 4 and 2 to get 8.

$$8-12\sqrt{2}\sqrt{3}+9\left(\sqrt{3}\right)^{2}$$

To multiply \sqrt{2} and \sqrt{3}, multiply the numbers under the square root.

$$8-12\sqrt{6}+9\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$8-12\sqrt{6}+9\times 3$$

Multiply 9 and 3 to get 27.

$$8-12\sqrt{6}+27$$

Add 8 and 27 to get 35.

$$35-12\sqrt{6}$$